Min value of Integer List when it can be empty in scala

Question

I am trying to find min value of integer list also when it can be empty.

scala> val minValue:Int = List() match {
     |   case Nil => -1
     |   case xs  => xs.min
     | }
<console>:9: error: diverging implicit expansion for type Ordering[B]
starting with method Tuple9 in object Ordering
     case xs => xs.min

though it worked well not non empty list.

scala> val minValue = List(1,2) match {
     |   case Nil => -1
     |   case xs  => xs.min
     | }
minValue: Int = 1

How do I find min with out sorting as one way I could think of is sort and get head with getOrElse defaultValue.

Answer 1

The problem is that the typechecker doesn't know what type to use for List(). If you specifically annotate the empty list with a type it should work just fine:

val minValue:Int = List.empty[Int] match { // ...

After discussing with @senia I realized you're probably using -1 as a "hack" to say "no minimum" rather than as an actual value. If that's the case, then using the Option type in Scala might be more clear, as you can return the Some variant for actual results and the None variant in the case of an empty list. Scala sequences actually already have a nice method for doing this for you:

scala> List.empty[Int].reduceLeftOption(_ min _)
res0: Option[Int] = None

scala> List(5, 2, 1, 3, 4).reduceLeftOption(_ min _)
res1: Option[Int] = Some(1)